30
Jul 19

Properties of root subspaces




Properties of root subspaces

Let A be a square matrix and let \lambda \in \sigma (A) be its eigenvalue. As we know, the nonzero elements of the null space N(A-\lambda I)=\{x:(A-\lambda I)x=0\} are the corresponding eigenvectors. This definition is generalized as follows.

Definition 1. The subspaces N_{\lambda }^{(k)}=N((A-\lambda I)^{k}), k=1,2,... are called root subspaces of A corresponding to \lambda .

Exercise 1. a) Root subspaces are increasing:

(1) N_{\lambda }^{(k)}\subset N_{\lambda }^{(k+1)} for all k\geq 1

and b) there is such p\leq n that all inclusions (1) are strict for k<p and

(2) N_{\lambda }^{(p)}=N_{\lambda }^{(p+1)}=...

Proof. a) If x\in N_{\lambda }^{(k)} for some k, then (A-\lambda I)^{k+1}x=(A-\lambda I)(A-\lambda I)^{k}x=0, which shows that x\in N_{\lambda }^{(k+1)}.

b) (1) implies \dim N_{\lambda }^{(k)}\leq \dim N_{\lambda }^{(k+1)}. Since all root subspaces are contained in C^{n}, there are k such that N_{\lambda }^{(k)}=N_{\lambda }^{(k+1)}. Let p be the smallest such k. Then all inclusions (1) are strict for k<p.

Suppose N_{\lambda}^{(k+1)}\setminus N_{\lambda }^{(k)}\neq \varnothing for some k\ge p. Then there exists x\in N_{\lambda }^{(k+1)} such that x\notin N_{\lambda}^{(k)}, that is, (A-\lambda I)^{k+1}x=0, (A-\lambda I)^{k}x\neq 0. Put y=(A-\lambda I)^{k-p}x. Then (A-\lambda I)^{p+1}y=(A-\lambda I)^{k+1}x=0, (A-\lambda I)^{p}y=(A-\lambda I)^{k}x\notin 0. This means
that y\in N_{\lambda }^{(p+1)}\setminus N_{\lambda }^{(p)} which contradicts the definition of p.

Definition 2. Property (2) can be called stabilization. The number p from (2) is called a height of the eigenvalue \lambda.

Exercise 2. Let \lambda \in \sigma (A) and let p be the number from Exercise 1. Then

(3) C^{n}=N_{\lambda }^{(p)}\dotplus \text{Img}[(A-\lambda I)^{p}].

Proof. By the rank-nullity theorem applied to (A-\lambda I)^{p} we have n=\dim N_{\lambda }^{(p)}+\dim \text{Img}[(A-\lambda I)^{p}]. By Exercise 3, to prove (3) it is sufficient to establish that L\equiv N_{\lambda}^{(p)}\cap \text{Img}[(A-\lambda I)^{p}]=\{0\}. Let's assume that L contains a nonzero vector x. Then we have x=(A-\lambda I)^{p}y for some y. We obtain two facts:

(A-\lambda I)^{p}y\neq 0 \Longrightarrow y\notin N_{\lambda }^{(p)},

(A-\lambda I)^{2p}y=(A-\lambda I)^{p}(A-\lambda I)^{p}y=(A-\lambda I)^{p}x=0\Longrightarrow y\in N_{\lambda }^{(2p)}.

It follows that y is a nonzero element of N_{\lambda }^{(2p)}\setminus N_{\lambda }^{(p)}. This contradicts (2). Hence, the assumption L\neq \{0\} is wrong, and (3) follows.

Exercise 3. Both subspaces at the right of (3) are invariant with respect to A.

Proof. If x\in N_{\lambda }^{(p)}, then by commutativity of A and A-\lambda I we have (A-\lambda I)^{p}Ax=A(A-\lambda I)^{p}x=0, so Ax\in N_{\lambda }^{(p)}.

Suppose x\in \text{Img}[(A-\lambda I)^{p}], so that x=(A-\lambda I)^{p}y for some y. Then Ax=(A-\lambda I)^{p}Ay\in \text{Img}[(A-\lambda I)^{p}].

Exercise 3 means that, for the purpose of further analyzing A, we can consider its restrictions onto N_{\lambda }^{(p)} and \text{Img}[(A-\lambda I)^{p}].

Exercise 4. The restriction of A onto N_{\lambda }^{(p)} does not have eigenvalues other than \lambda .

Proof. Suppose Ax=\mu x, x\neq 0, for some \mu . Since x\in N_{\lambda }^{(p)}, we have (A-\lambda I)^{p}x=0. Then (A-\lambda I)x=(\mu -\lambda )x and 0=(A-\lambda I)^{p}x=(\mu -\lambda )^{p}x. This implies \mu =\lambda .

Exercise 5. The restriction of A onto \text{Img}[(A-\lambda I)^{p}] does not have \lambda as an eigenvalue (so that A-\lambda I is invertible).

Proof. Suppose x\in \text{Img}[(A-\lambda I)^{p}] and Ax=\lambda x, x\neq 0. Then x=(A-\lambda I)^{p}y for some y\neq 0 and 0=(A-\lambda  I)x=(A-\lambda I)^{p+1}y. By Exercise 1 y\in N_{\lambda }^{(p+1)}=N_{\lambda }^{(p)} and x=(A-\lambda I)^{p}y=0. This contradicts the choice of x.

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